Thanks for digging up the drag equation, but lets look at it as a percentage.
Since we are comparing only temperature differences to our drag, we can simplify EVERYTHING down to only a ratio of the air densities (rho). [That's because we want drag differences using the same car (Cd and A) at the same velocity.]
Rho1/Rho2 = percent drag
Unfortunately, calculating Rho isn't as easy but is necessary for both the drag equation and the equation I provided above. Using the Ideal Gas Equation of
Rho = P / (R*T) where P is air pressure (Pascals), R is a constant 287.05 (Joules/(Kilogram*Kelvin)), and T is temperature (in Kelvin).
You can get temperature and pressure from the weather channel or Weather.com. **However, this pressure is not the observed pressure but a converted pressure to compare to standard day sea level pressure. To convert it back to an observed pressure (that's what you want), use the following simplified equation which assumes constant temperature lapse rate and gravity:
Po = P*(1-0.0000225577*h)^5.257576
Where h is the elevation of the observation, Po is the observed pressure, and P is the pressure given from the weather service. You can easily get this figure from the Google search page by typing "[local airport] elevation". All units are metric. Convert in-hg to Pascals (Google search an online calculator).
Here is a live example for you.
In Huntsville, AL (where I live) Weather.Com says the temperature is 41 F and the pressure is 30.09 in-Hg. Let's teleport to Honolulu, HI. The temp there is 67F and the pressure is 30.06 in-Hg.
Calculate Observed Pressure!
Huntsville --- 41F -> 278.15K
, 30.09 in-Hg -> 99597 Pa
Honolulu --- 67F -> 292.59K
, 30.06 in-Hg -> 101746 Pa
Plug-in Ideal Gas Equation for each City:
Huntsville air density --- 99597/(287.05*278.15) = 1.2474 kg/m^3
Honolulu air density --- 101746/(287.05*292.59) = 1.2114 kg/m^3
1.2474/1.2114 = 1.02969 or 102.97% or 2.97% increase in aerodynamic drag in Huntsville...
Which is just what I need... another excuse to move to Hawaii